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Problem on probability

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Problem on probability

by Pinku » Wed Jul 28, 2010 4:24 pm
Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously
and at random from the box. What is the probability that neither of the lightbulbs selected will be
one that is defective?
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by Patrick_GMATFix » Wed Jul 28, 2010 4:59 pm
We want to find out how to pick 2 good light bulbs from a batch that contains 2 bad bulbs and 18 good ones. 2 ways to do this.

Solution A

Probability is always the ratio (desirable outcomes)/(all possible outcomes). One way to solve this question is to find each part of the ratio.

Since there are only 2 defective bulbs, there must be 18 good light bulbs. The number of ways to select 2 good light bulbs equals the number of ways to choose 2 from 18 available. This is a combination of 2 out of 18, or 18!/[(2!)(16!)] = 153 desirable outcomes.

The total number of outcomes possible is equivalent to the number of pairs of bulbs that can be selected from 20 bulbs available. This is a combination. How many ways can you pick 2 from 20? This is 20!/[(2!)(18!)] = 190 possible selections.

So the probability we want is (desirable outcomes)/(all possible outcomes) >> 153/190.


Solution B

A 2nd approach is to rely on the rule of probability that states that the probability that two events A and B happen is the product of the individual probabilities. P(A & B) = P(A)*P(B). In this case we want to find out the probability that a good bulb is picked twice.

The probability that the first bulb picked is good is 18/20 because there are 18 good out of 20 possible picks. The probability that the 2nd bulb picked is also good is 17/19 because after the first is picked, only 17 good bulb remains out of 19 available. The probability that both events happen is (18/20)(17/19)= 153/190

Notes
* Whether the bulbs are picked simultaneously or in succession has no bearing on the probability. This is a red herring.

* When using the 2nd approach, always assume that all events are occurring as desired. This is why the 2nd probability in our calculation is 17/19: we assume that a good bulb was picked on the first selection.

* If you have trouble with probabilities, generate timed drills of similar questions by setting topic='Combinatorics' and difficulty='600-700 & 700+' in the Drill Generator

Good luck,
-Patrick
Last edited by Patrick_GMATFix on Wed Jul 28, 2010 5:36 pm, edited 1 time in total.
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by Rahul@gurome » Wed Jul 28, 2010 5:07 pm
Pinku wrote:Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously
and at random from the box. What is the probability that neither of the lightbulbs selected will be
one that is defective?
Out of 20 bulbs, 2 are defective and 18 are good.
So, probability of selecting one good bulb = 18/20
Probability of selecting the 2nd good bulb = 17/19

So, required probability = (18/20)(17/19) = 153/190
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by Patrick_GMATFix » Wed Jul 28, 2010 5:21 pm
oops. I solved for probability that BOTH light bulbs be defective. I'll have to edit my solution.... (solution now fixed)
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by Patrick_GMATFix » Wed Jul 28, 2010 9:02 pm
As I wrote in my notes, from a mathematical standpoint, whether the bulbs are picked simultaneously or in succession does not affect probability.

2/20 is the probability that 1 bad bulb is picked at the start of the experiment.
9/10 is the probability that 1 good bulb is picked at the start of the experiment.

-Patrick
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